Question

plz help me
find x And y

here x and y both belongs to integer
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615+(x)^2=2^y

Answer 1

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First, x must be odd.

as ,615 is odd and${2}^{y}$=even and odd+odd=even but odd+ even=odd and y>9 and square of an odd have unit digits like 1,9,5,9

So,${615}+{x}^{2}$ have unit digits 6,4,0.

So, ${2}^{y}$ have unit digits 4 & 6 .

so,

y is even=2k

so, $615+{x}^{2}={2}^{y}$

$615=({2}^{k}+x) ({2}^{k}-x)$

So,we must split 615 as product of two natural number.

Then ,${615}={41} * {15}={123} * {5}={205} *{3}$

Now,

Let x and y positive integer then only acceptable value is

${2}^{k}+x=123…........…i$

and,

${2}^{k}-x=5...........……ii$

Now,

(i) + (ii)

${2}^{(k+1)}=128$

k=6

y=12

(i) - (ii)

2x=118

x=59

So,x=59 & y=12

If x negative integer then

${2}^{k}+x=5……...........i$

and,

${2}^{k}-x=123…...........ii$

(i) + (ii)

${2}^{(k+1)}=128$

k=6

y=12

(i) - (ii)

2x=-118

x=-59

So,x=-59 & y=12

So,y=12 & x=59,-59.

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